# Homework Help: Hanging Cable

1. Jul 14, 2004

### e(ho0n3

A uniform steel cable of weight mg is suspended between two equal elevation points. Determine the tension in the cable (a) at its lowest point, and (b) the points of attachment.

Judging from the problem, it seems that the tension in (a) and (b) is different. Is this necessarily so? If it is, why? Do I have to calculate the tensions to determine this?

2. Jul 14, 2004

### Staff: Mentor

Think about it. At any point, the vertical component of tension must support the weight of the cable below that point. So, which point has more cable to support?

3. Jul 14, 2004

### rayjohn01

Catenary

luckilly they did not ask to prove it's shape. This is not difficult if you remember that for the system as a whole the horizontal forces will equal zero and the same for vertical, start with the fact that the weight ( downward force) must be matched by the suspension. Then think of the fact that at the lowest point there can be no vertical forces.
Ray.

4. Jul 14, 2004

### e(ho0n3

Are you saying then that the tension at the points of attachment is greater than the tension at the lowest point of the cable? I just find it difficult to digest that a rope in equilibrium can have varying tensions at different points.

5. Jul 14, 2004

### e(ho0n3

I figured this already. However, if there are no vertical forces on the lowest point of the rope, then the tension is purely horizontal. So at this point, there is a tension vector pointing to the left, and another to the right. These two cancel. I don't know how to proceed. The shape of the rope looks like a graph of cosh by the way.

6. Jul 14, 2004

### rayjohn01

not sure

to e -- yes they cancel but they are also equal to two OTHER horizontal forces if you can find them.
The shape is a ( form of ) cosh function ( can't remember precisely ) it took a long time ( centuaries ago) to distinguish it from eliptical.
Don't listen to much to me I also get confused by mechanical problems , but ( and I'm thinking allowed) if all points are at equilibrium and there is no motion it seems that at any slice through the cable the vertical forces and the horizontal forces must each cancel. The rope or chain tension is the vector sum of these and of course the angle changes throughout the rope.
Another point you did not state any dimensions but if I attempt to pull the rope through one of the suspension points so as to reduce it's sag ( especially close to zero) the tension must head towards infinity , which suggests you must account for rope length compared to suspension separation -- that sounds like a nice calculus problem to me.

Last edited: Jul 14, 2004
7. Jul 15, 2004

### Staff: Mentor

tension varies along that rope!

Absolutely!
That's because you are probably used to problems that allow the "massless rope" approximation. A section of massless rope will have the same tension throughout.

Imagine a heavy chain hanging from the ceiling. It's certainly in equilibrium. Take the chain as a whole: the tension at the top must equal the weight of the entire chain. Now just look at the bottom half of the chain: the tension at the half-way point is the only upward force on that bottom section of chain--that tension must therefore equal the weight of half the chain.

Meditate on this long enough and it will become obvious.

8. Jul 15, 2004

### Staff: Mentor

tension cancels? What does that mean?

Right. At the lowest point, the force produced by the rope's tension is horizontal.
I don't know what you mean by "cancel". At any point the tension in the rope pulls on on both sides of the rope at that point. These forces don't "cancel", since they act on different sections of rope.
If you can figure out the angle that the rope makes at the attachment points, then you can figure out the horizontal and and vertical components of the force on the rope at that point. Apply equilibrium constraints.

9. Jul 15, 2004

### e(ho0n3

Yes indeed.

OK, I'm convinced.

Let T be the tension. What is the sum of the forces in the horizontal direction at the lowest point on the rope? Wouldn't that be T - T = 0 (this is what I mean by "cancel")?

I'm given the angle already, but I can only figure out the tension at the points of attachment (since they don't "cancel" each other).

10. Jul 15, 2004

### Staff: Mentor

The sum of the forces on what? The 0-length piece of rope exactly in the middle? By your logic, tension always cancels out--since the same logic applies anywhere on the rope.
It's easy to show that the horizontal component of the tension force is the same throughout the rope. For example, analyze the right half of the rope: the tension at the bottom pulling left must balance the tension at the attachment pulling right (which is the horizontal component of the tension at the attachment).

If you're given the angle, it's a breeze.

11. Jul 15, 2004

### e(ho0n3

Excellent point. This never came to mind. I guess that does it for this problem. Thanks.